If a line intersects a curve in $n-1$ rational points is the $n$'th point also rational?

Question

I am trying to solve problem 5.40 from Fulton's book on algebraic curves, stating the following

Let $\mathbb{K}_0$ be a subfield of $\mathbb{K}$, and $\mathbb{K}$ algebraically closed. Suppose a nonsingular cubic $C$ is rational over $\mathbb{K_0}$ (all the coefficients defining $C$ are in $\mathbb{K}_0$). Let $C(\mathbb{K}_0)$ be the set of points of $C$ that are rational over $\mathbb{K}_0$. Then, if $P,Q\in C(\mathbb{K}_0)$, and $L$ the line passing through $P$ and $Q$ intersects $C$ in a third point $R$, show that $R\in C(\mathbb{K}_0)$.

Even though in most of the book Fulton supposes that $\mathop{Char}(\mathbb{K})=0$, I am not sure if it is necessary in order to solve this problem nor if we need it for Bezout's theorem, but I believe it's not. By this point in the book Bezout's and Noether's theorem are already covered.

I know a way to solve the problem by putting the curve in Weierstrass form and explicitly computing $R$, but I don't think it works in charactersistics $2$ and $3$, and the wording of the problem suggests that it can be proved without resorting to an explicit construction of $R$.

Furthermore, is this result true in general? If a line $L$ intersects a degree $n$ rational curve $F$ in $n-1$ rational points, then the $n$'th point of intersection is also rational.

If $L$ intersects $F$ in $n-2$ rational points this is not true. We can see this by setting $\mathbb{K}=\mathbb{C}$, $\mathbb{K}_0=\mathbb{R}$, $F=x^4-y$ and $L=y-1$. Then we have $$ L\cdot_\mathbb{R} F = (1,1) + (-1,1), $$ while $$ L\cdot_\mathbb{C} F = (1,1) + (-1,1) + (i,1) + (-i,1). $$

Answer 1

I don't know if it is true if $F \subset \overline{F}$ is inseparable, but in the case that it is separable: the Galois group $G = Gal(\overline{F}/F)$ acts on the intersection points by acting on the coordinates, so if the $n-1$ points are rational (which is equivalent to being fixed by $G$), then the $n$th point is also fixed by $G$, so rational.

In the case of your example, $G = \mathbb{Z}/2\mathbb{Z}$ acts your four points by transposing $(i,1)$ and $(-i,1)$ and fixing $(1,1)$ and $(-1,1)$.

Answer 2

Consider a curve defined over a field $F$ by a polynomial equation $f(x,y)=0$ of degree $n$, and a line defined by a linear equation $y=mx+c$, again with coefficients in $F$. Over the algebraic closure of $F$ they definitely will have $n$ intersections, counting multiplicity, of course.

However if $n-1$ of the intersections are defined over $F$ then the $n$th one is also defined over $F$. This is because sum of the $x$-coordinates of the $n$ intersection points is the negative of the coefficient $x^{n-1}$ in $f(x, mx+c)=0$.